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How do you find all the zeros of $f (x) = x^{2} - 4 x - 60$ $f(x) = x^2 - 4x - 60$ with its multiplicities?

1 Answer

Apr 6, 2016

$f (x)$ $f(x)$ has zeros at $x = 6$ $x=6$ and $x = - 4$ $x=-4$ (there are only $2$ $2$ zeros including multiplicity)

Explanation:

Factoring
$XXX f (x) = x^{2} - 4 x - 60$ $color(white)("XXX")f(x)=x^2-4x-60$

$XXXXX = (x + 4) (x - 6)$ $color(white)("XXXXX")=(x+4)(x-6)$

When $f (x) = 0$ $f(x)=0$
either
$XXX (x + 4) = 0 \to x = - 4$ $color(white)("XXX")(x+4)=0 rarr x=-4$
or
$XXX (x - 6) = 0 \to x = + 6$ $color(white)("XXX")(x-6)=0 rarr x=+6$

Since $f (x)$ $f(x)$ is a quadratic (i.e. of degree $2$ $2$ )
it has (including multiplicity and possible complex values) exactly $2$ $2$ zeroes.

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